Question 4
∫1_(𝑎 + 𝑐)^(𝑏 + 𝑐)▒〖 𝑓(𝑥) 𝑑𝑥〗 is equal to
∫1_(𝑎 )^(𝑏 )▒〖𝑓(𝑥−𝑐) 𝑑𝑥〗 (B) ∫1_(𝑎 )^(𝑏 )▒〖𝑓(𝑥+𝑐) 𝑑𝑥〗
(C) ∫1_(𝑎 )^(𝑏 )▒〖𝑓(𝑥) 𝑑𝑥〗 (D) ∫1_(𝑎 −𝑐)^(𝑏−𝑐 )▒〖𝑓(𝑥) 𝑑𝑥〗
∫1_(𝑎 + 𝑐)^(𝑏 + 𝑐)▒〖 𝑓(𝑥) 𝑑𝑥〗
Putting 𝒙=𝒕+𝒄
Differentiating w.r.t. 𝑥
𝑑𝑥=𝑑𝑡
Now,
when 𝒙 varies from a + c to b + c
then 𝒕 varies from a to b
Therefore
∫1_(𝑎 + 𝑐)^(𝑏 + 𝑐)▒〖 𝑓(𝑥) 𝑑𝑥〗 =∫_𝑎^𝑏▒𝑓(𝑡+𝑐)𝑑𝑡
Changing variables – using Property 1
=∫_𝒂^𝒃▒𝒇(𝒙+𝒄)𝒅𝒙
So, the correct answer is (b)
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